3.345 \(\int \frac{a+b \sinh ^{-1}(\frac{\sqrt{1-c x}}{\sqrt{1+c x}})}{1-c^2 x^2} \, dx\)

Optimal. Leaf size=133 \[ \frac{b \text{PolyLog}\left (2,e^{-2 \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right )}{2 c}-\frac{\left (a+b \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^2}{2 b c}-\frac{\log \left (1-e^{-2 \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right ) \left (a+b \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )}{c} \]

[Out]

-(a + b*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2/(2*b*c) - ((a + b*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*Log[1
- E^(-2*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])])/c + (b*PolyLog[2, E^(-2*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])]
)/(2*c)

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Rubi [A]  time = 0.121832, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 7, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.184, Rules used = {206, 6681, 5659, 3716, 2190, 2279, 2391} \[ -\frac{b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right )}{2 c}+\frac{\left (a+b \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^2}{2 b c}-\frac{\log \left (1-e^{2 \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right ) \left (a+b \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )}{c} \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])/(1 - c^2*x^2),x]

[Out]

(a + b*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2/(2*b*c) - ((a + b*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*Log[1 -
E^(2*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])])/c - (b*PolyLog[2, E^(2*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])])/(
2*c)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 6681

Int[((a_.) + (b_.)*(F_)[((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_.)*(x_)]])^(n_.)/((A_.) + (C_.)*(x_)^
2), x_Symbol] :> Dist[(2*e*g)/(C*(e*f - d*g)), Subst[Int[(a + b*F[c*x])^n/x, x], x, Sqrt[d + e*x]/Sqrt[f + g*x
]], x] /; FreeQ[{a, b, c, d, e, f, g, A, C, F}, x] && EqQ[C*d*f - A*e*g, 0] && EqQ[e*f + d*g, 0] && IGtQ[n, 0]

Rule 5659

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Tanh[x], x], x, ArcSinh
[c*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c
+ d*x)^(m + 1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{a+b \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}{1-c^2 x^2} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{a+b \sinh ^{-1}(x)}{x} \, dx,x,\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}{c}\\ &=-\frac{\operatorname{Subst}\left (\int (a+b x) \coth (x) \, dx,x,\sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )}{c}\\ &=\frac{\left (a+b \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2}{2 b c}+\frac{2 \operatorname{Subst}\left (\int \frac{e^{2 x} (a+b x)}{1-e^{2 x}} \, dx,x,\sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )}{c}\\ &=\frac{\left (a+b \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2}{2 b c}-\frac{\left (a+b \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{c}+\frac{b \operatorname{Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )}{c}\\ &=\frac{\left (a+b \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2}{2 b c}-\frac{\left (a+b \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{c}+\frac{b \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{2 c}\\ &=\frac{\left (a+b \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2}{2 b c}-\frac{\left (a+b \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{c}-\frac{b \text{Li}_2\left (e^{2 \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{2 c}\\ \end{align*}

Mathematica [A]  time = 0.0279931, size = 127, normalized size = 0.95 \[ \frac{\left (a+b \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right ) \left (a+b \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )-2 b \log \left (1-e^{2 \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right )\right )-b^2 \text{PolyLog}\left (2,e^{2 \sinh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right )}{2 b c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])/(1 - c^2*x^2),x]

[Out]

((a + b*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*(a + b*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]] - 2*b*Log[1 - E^(2*A
rcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])]) - b^2*PolyLog[2, E^(2*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])])/(2*b*c)

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Maple [A]  time = 0.006, size = 263, normalized size = 2. \begin{align*} -{\frac{a\ln \left ( cx-1 \right ) }{2\,c}}+{\frac{a\ln \left ( cx+1 \right ) }{2\,c}}+{\frac{b}{2\,c} \left ({\it Arcsinh} \left ({\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}} \right ) \right ) ^{2}}-{\frac{b}{c}{\it Arcsinh} \left ({\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}} \right ) \ln \left ( 1+{\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}}+\sqrt{1+{\frac{-cx+1}{cx+1}}} \right ) }-{\frac{b}{c}{\it polylog} \left ( 2,-{\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}}-\sqrt{1+{\frac{-cx+1}{cx+1}}} \right ) }-{\frac{b}{c}{\it Arcsinh} \left ({\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}} \right ) \ln \left ( 1-{\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}}-\sqrt{1+{\frac{-cx+1}{cx+1}}} \right ) }-{\frac{b}{c}{\it polylog} \left ( 2,{\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}}+\sqrt{1+{\frac{-cx+1}{cx+1}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2)))/(-c^2*x^2+1),x)

[Out]

-1/2*a/c*ln(c*x-1)+1/2*a/c*ln(c*x+1)+1/2*b/c*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2-b/c*arcsinh((-c*x+1)^(1/2
)/(c*x+1)^(1/2))*ln(1+(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1+(-c*x+1)/(c*x+1))^(1/2))-b/c*polylog(2,-(-c*x+1)^(1/2)/(
c*x+1)^(1/2)-(1+(-c*x+1)/(c*x+1))^(1/2))-b/c*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2))*ln(1-(-c*x+1)^(1/2)/(c*x+1)
^(1/2)-(1+(-c*x+1)/(c*x+1))^(1/2))-b/c*polylog(2,(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1+(-c*x+1)/(c*x+1))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{8} \, b{\left (\frac{2 \,{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )} \log \left (c x + 1\right ) - \log \left (c x + 1\right )^{2} + 2 \, \log \left (c x + 1\right ) \log \left (-c x + 1\right ) - \log \left (-c x + 1\right )^{2} - 4 \,{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )} \log \left (\sqrt{2} + \sqrt{-c x + 1}\right )}{c} + 8 \, \int -\frac{\sqrt{2} \log \left (c x + 1\right ) - \sqrt{2} \log \left (-c x + 1\right )}{4 \,{\left (\sqrt{2} c x +{\left (c x - 1\right )} \sqrt{-c x + 1} - \sqrt{2}\right )}}\,{d x}\right )} + \frac{1}{2} \, a{\left (\frac{\log \left (c x + 1\right )}{c} - \frac{\log \left (c x - 1\right )}{c}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2)))/(-c^2*x^2+1),x, algorithm="maxima")

[Out]

-1/8*b*((2*(log(c*x + 1) - log(-c*x + 1))*log(c*x + 1) - log(c*x + 1)^2 + 2*log(c*x + 1)*log(-c*x + 1) - log(-
c*x + 1)^2 - 4*(log(c*x + 1) - log(-c*x + 1))*log(sqrt(2) + sqrt(-c*x + 1)))/c + 8*integrate(-1/4*(sqrt(2)*log
(c*x + 1) - sqrt(2)*log(-c*x + 1))/(sqrt(2)*c*x + (c*x - 1)*sqrt(-c*x + 1) - sqrt(2)), x)) + 1/2*a*(log(c*x +
1)/c - log(c*x - 1)/c)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{b \operatorname{arsinh}\left (\frac{\sqrt{-c x + 1}}{\sqrt{c x + 1}}\right ) + a}{c^{2} x^{2} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2)))/(-c^2*x^2+1),x, algorithm="fricas")

[Out]

integral(-(b*arcsinh(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a)/(c^2*x^2 - 1), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh((-c*x+1)**(1/2)/(c*x+1)**(1/2)))/(-c**2*x**2+1),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{b \operatorname{arsinh}\left (\frac{\sqrt{-c x + 1}}{\sqrt{c x + 1}}\right ) + a}{c^{2} x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2)))/(-c^2*x^2+1),x, algorithm="giac")

[Out]

integrate(-(b*arcsinh(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a)/(c^2*x^2 - 1), x)